Question

 

For the reaction below,

A. determine the order of each reactant (3 points)

B. the overall order (1 point)

C. the average rate constant (1 points)

using the method of initial rates.

2A + B +C à D + 2E

Experiment	[A]	[B]	[C]	Initial Rate (M/s)
1	1.5	0.50	2.0	5.0 x 10-3
2	0.75	0.50	1.0	2.5 x 10-3
3	0.75	0.50	2.0	5.0 x 10-3
4	1.5	1.00	1.0	2.0 x 10-2
5	1.5	0.50	1.0	2.5 x 10-3

Answer 1

In order to calculate the rate law expression for a A+B reaction, we need to apply Initial Rates Method.

Note that the generic formula goes as follows:

r = k [A]^a [B]^b

Note that if we got at least 3 sets of point, in which we have A and B constant, then we could use:

r1 / r2 = (k1 [A]1^a [B]1^b) / (k2 [A]2^a [B]2^b)

If we assume K1 and K2 are constant, then K1= K2 cancel each other

r1 / r2 = ([A]1^a [B]1^b) / ( [A]2^a [B]2^b)

Then, order according to [A] and [B]

r1 / r2 = ([A]1/[A2])^a * ([B]1/[B]2)^b

If we get two points in which A1 = A2, then we could get B, and vise versa for A...

From the data shown in YOUR table

choose point 1 and 3 so b and c cancel

(5*10^-3)/(5*10^-3) = (1.5/0.75)^a

1 = 2^a

a = 0

now, choose point2 and 3 so a and b cancel

(2.5*10^-3)/(5*10^-3) = (0.75/0.75)^a * (0.5/0.5)^b * (1/2)^c

0.5 = 0.5^c

c = 1

now, choose points: 4 and 5

(2*10^-2)/(2.5*10^-3) = (1.5/1.5)^a * (1/0.5)^b * (1/1)^c

8 = 2 ^b

ln(8)/ln(2) = b

b = 3

then

Order --> Rate = k * [A]^0 * [B]^3 *[C]^1

b)

overal order

addition of orders

n = a+b+c = 0+3+1 = 4th order

c)

find k

choose any point

Rate = k*[A]^0 [B]^3 [C]

(5*10^-3) = k (1.5)^0 * (0.5)^3 * (2)

k = (5*10^-3) / ((0.5)^3 * (2))

k = 0.02